Week 7 MATH 225N Hypothesis Testing Questions and Answers 2021_Already Graded A.

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Week 7 Hypothesis Testing Q & A 1. Steve listens to his favorite streaming music service when he works out. He wonders whether the service algorithm does a good job of finding random songs that he w... ill like more often than not. To test this, he listens to 50 songs chosen by the service at random and finds that he likes 32 of them. Use Excel to test whether Steve will like a randomly selected song more than not and then draw a conclusion in the context of a problem. Use α = 0.05. Typeequationhere . Ho: p = ≤ 0.5 (50%) p = 0.5 Ha: p = > 0.5 (strictly ¿≠ ) P-value = 0.02 which is < α=0.05 we reject Ho and support the Ha Hypothesis Test for p population proportion Level of Significance 0.05 (decimal ) Proportion under H0 0.5000 (decimal ) n 50 Number of Successes 32 Sample Proportion 0.64000 0 StDev 0.50000 0 SE 0.07071 1 Test Statistic (z) 1.97989 9 One-Sided p-value 0.02385 2 Two-Sided p-value 0.04770 4 Right-Tailed (>) 1.644854 Left-Tailed (<) -1.644854 Two-Tailed (≠) ± 1.959964 Answer: Reject the null hypothesis. There is sufficient evidence to prove that Steve will like a random selected song more often than not. 2. A magazine regularly tested products and gave the reviews to its customers. In one of its reviews, it tested 2 types of batteries and claimed that the batteries from company A outperformed batteries from company B in 108 of the tests. There were 200 tests. Company B decided to sue the magazine, claiming that the results were not significantly different from 50% and that the magazine was slandering its good name.Use Excel to test whether the true proportion of times that Company A’s batteries outperformed Company B’s batteries is different from 0.5. Identify the p=value rounding it to 3 decimal places. Ho: p = 0.5 Ha ≠ 0.5 (two tailed test) n = 200 (α is not given soleave it 0.05) Hypothesis Test for p population proportion Level of Significance 0.05 Proportion under H0 0.5000 n 200 Number of Successes 108 Sample Proportion 0.540000 StDev 0.500000 SE 0.035355 Test Statistic (z) 1.131371 One-Sided p-value 0.129238 Two-Sided p-value 0.258476 Right-Tailed (>) 1.644854 Left-Tailed (<) - 1.644854 Two-Tailed (≠) ± 1.959964 Answer: 0.258 (because it is a two tailed test). We are not rejecting the null hypothesis and we do not have evidence to support the alternative hypothesis. 3. A candidate in an election lost by 5.8% of the vote. The candidate sued the state and said that more than 5.8% of the ballots were defective and not counted by the voting machine, so a full recount would need to be done. His opponent wanted to ask for the case to be dismissed, so she had a government official from the state randomly select 500 ballots and count how many were defective. The official found 21 defective ballots. Use Excel to test if the candidates claim is true and that < 5.8% of the ballots were defective. Identify the p=value rounding to 3 decimal places. Ho: p = ≥ 0.058 Ha ¿0.058 (one tailed test) n = 500 (α is not given soleave it 0.05) Hypothesis Test for p population proportion Level of Significance 0.05 (decimal) Proportion under H0 0.0580 (decimal) n 500 Number of Successes 21 Sample Proportion 0.042000 StDev 0.233743 SE 0.010453 Test Statistic (z) -1.530613 One-Sided p-value 0.063008 Two-Sided p-value 0.126016 Right-Tailed (>) 1.644854 Left-Tailed (<) -1.644854 Two-Tailed (≠) ± 1.959964 Answer: 0.063 4. A researcher claims that the incidence of a certain type of cancer is < 5%. To test this claim, a random sample of 4000 people are checked and 170 are found to have the cancer. The following is the set up for the hypothesis: Ho = 0.05 Ha = < 0.05 In the example the p-value was determined to be 0.015. Come to a conclusion and interpret the results of this hypothesis test for a proportion (use a significance level of 5%) Answer: The decision is to reject the null hypothesis. The conclusion is that there is enough evidence to support the claim. 5. A researcher is investigating a government claim that the unemployment rate is < 5%. TO test this claim, a random sample of 1500 people is taken and it is determined that 61 people were unemployed. Ho: p = 0.05 Ha: p < 0.05 Find the p-value for this hypothesis test for a proportion & round to 3 decimal places. Hypothesis Test for p population proportion Level of Significance 0.05 Proportion under H0 0.0500 n 1500 Number of Successes 61 Sample Proportion 0.04066 7 StDev 0.21794 5 SE 0.00562 7 Test Statistic (z) - 1.658577 One-Sided p-value 0.04845 7 Two-Sided p-value 0.09691 4 Answer: 0.048 6. An economist claims that the proportion of people that plan to purchase a fully electric vehicle as their next car is greater than 65%. To test this claim, a random sample of 750 people were asked if they planned to purchase a fully electric vehicle as their next car. Of this 750, 513 indicated that they plan to purchase an electric vehicle. Ho: p = 0.65 Ha; p = >0.65 Find the p-value for this hypothesis test for a proportion & round to 3 decimal places. Hypothesis Test for p population proportion Level of Significance 0.05 Proportion under H0 0.6500 n 750 Number of Successes 513 Sample Proportion 0.68400 0 StDev 0.47697 0 SE 0.01741 6 Test Statistic (z) 1.95217 5 One-Sided p-value 0.02558 8 Two-Sided p-value 0.05117 6 Answer: 0.026 7. Colton makes the claim to his classmates that < 50% of newborn babies born this year in his state are boys. To prove this claim, he selects a random sample of 344 birth records in his state from this year. Colton found that 176 of the newborns were boys. What are the null and alternative hypothesis for this hypothesis test. Answer: Ho: 0.5Ha: <0.5 8. An Airline company claims that in its recent advertisement that at least 94% of passenger luggage that is lost is recovered and reunited with their customer within 1 day. Hunter is a graduate student studying statistics. For a research project, Hunter wants to find out whether there is sufficient evidence in support of the airline company’s claim. He randomly selects 315 passengers whose luggage was lost by the airlines and found out that 276 of those passengers were reunited with their luggage within 1 day. Are all of the conditions for his hypotheses test met, and if so, what are the Ho and Ha for this hypothesis test? For a binomial Model to follow the normal model, the following condition must be satisfied: Success count = n * p ≥ 5 and Failure count ≥5 Example: success count 315 * 0.94 = 296.1 and failure count 315-296.1 = 18.9 so it meets the conditions. Answer: All of the conditions were met and the Ho = 0.94; Ha = >0.94 9. A college administrator claims that the proportion of students who are nursing majors is > 40%. To test this claim, a group of 400 students are randomly selected and its determined that 190 are nursing majors. The following is the set up for the hypothesis test: Ho: p = .40 and Ha: p = >.40 Find the test statistics for this hypothesis test for a proportion & round to 2 decimal places. Answer: 3.06 Level of Significance 0.05 Proportion under H0 0.4000 n 400 Number of Successes 190 Sample Proportion 0.475000 StDev 0.489898 SE 0.024495 Test Statistic (z) 3.061862 One-Sided p-value 0.001107 Two-Sided p-value 0.002214 10. A hospital administrator claims that the proportion of knee surgeries that are successful are 87%. To test this claim, a random sample of 450 patients who underwent knee surgery is taken and it is determined that 371 patients had a successful knee surgery operation. Ho: p = 0.87 Ha: p ≠ 0.87 (two sided tail) Find the test statistics for this hypothesis test for a proportion & round to 2 decimal places. Answer: -2.87 (this would be rejected) Level of 0.05Significance Proportion under H0 0.8700 n 450 Number of Successes 371 Sample Proportion 0.824444 StDev 0.336303 SE 0.015853 Test Statistic (z) -2.873534 One-Sided p-value 0.002052 Two-Sided p-value 0.004104 11. Jose, a competitor in cup stacking, has a sample stacking time mean of 7.5 seconds from 13 trials. Jose still claims that his average stacking time is 8.5 seconds, and the low average can be contributed to chance. At the 2% significant level, does the data provide sufficient evidence to conclude that Jose’s mean stacking time is less than 8.5 seconds? Given the sample data below, select or reject the hypothesis. (If p=value is < alpha value, we would automatically reject the hypothesis) Ho: μ = 8.5 Ha: μ = <8.5 α = 0.02 (significance level) Zo = -2.18 P = 0.0146 Answer: Reject the null hypothesis because the p value 0.0146 is less than the significance level 0.02 12. Marty, a typist, claims his average typing speed is 72 wpm. During a practice session, Marty has a sample typing speed mean of 84 wpm based on 12 trials. At the 5% significance level, does the data provide sufficient evidence to conclude that his mean typing speed is >72 wpm? Accept or reject the hypothesis given the data below. Ho: μ=72wpm ; Ha: μ=¿72wpm ; α=0.05 (significance level) ; Zo = 2.1; p = 0.018 Answer: Reject the null hypothesis because the p-value 0.018 is less than the significance level α=0.05 13. What is the p-value of a right-tailed one mean hypothesis test, with a test statistic of Zo = 2.1? (Do not round your answer. Compute your answer using a value from the table. (Value in table was 0.982) 1 – 0.982 = p=value of 0.018 Answer: 0.018 14. What is the p-value of a two-tailed one mean hypothesis test, with a test statistic of Zo = 0.27? (Do not round your answer. Compute your answer using a value from the table. (Value in table was 0.606)1-0.606 = P-value of 0.394 because it is two tailed, you multiply 2 x .394 = .788 Answer: 0.788 15. Raymond, a typist, claims his average typing speed is 89 wmp. During a practice session, Raymond has a sample typing speed mean of 95.5 wmp based on 15 trials. At the 1% significance level, does the data provide sufficient evidence to conclude that his mean typing speed is > 89 wmp? Accept or reject the hypothesis given the sample data below: Ho: μ=89 ; Ha: μ=¿89 wmp α=0.02 Sig Level; Zo = 2.75; p = 0.003 Answer: Reject the null hypothesis because p-value 0.003 is less than alpha 0.02 16. Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61 employees who work in Yarmouth and record their annual salary. Gina calculates the sample mean income to be $56.500 per year with a sample standard deviation of $3750. Using the alternative hypothesis, Ha = μ=¿ $55,000 , find the test statistic τ and the p-value for the appropriate hypothesis test. Round the τ to 2 decimal places and the p-value to 3 decimal places. Hypothesis Test for µ for unknown population stdev Level of Significance 0.1 (decimal ) Mean under H0 55000 n 61 Sample Mean 56500 StDev 3750 SE 480.13830 0 Test statistic (t) 3.124100 One-Sided p-value 0.001373 Two-Sided p-value 0.002746 Answer: t = 3.12; p-value = 0.001 17.A college administrator claims that the proportion of students that are nursing majors is less than 40%. To test this claim, a group of 400 students are randomly selected and its determined that 149 are nursing majors. The following is the setup for this hypothesis test: H0:p=0.40 Ha:p<0.40 Find the p-value for this hypothesis test and round your answer to 3 decimal places.Hypothesis Test for p population proportion Level of Significance 0.05 Proportion under H0 0.4000 n 400 Number of Successes 149 Sample Proportion 0.37250 [Show More]

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